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I understand that the inverse of a bijection is a bijection. To proof this you need to formally proof it is injective and surjective. I can prove it is injective and i understand what it is in the following way:

Let $f:A\to B$ be a bijection and let $f^{-1}: B\to A$ be its inverse. To show $f^{-1}$ is a bijection, you must show it is an injection: Let $x_1,x_2 \in B$ such that $f^{-1}(x_1)=f^{-1}(x_2)$. The the inverse we have $x_1= f(f^{-1}(x_2))=x_2$. This shows $f^{-1}$ is injective.

I am having a hard time proving this is surjective formally. I understand this is the definition of a surjection but I dont understand how this applies. Basically I dont understand it.

$$\forall b_i \in B~~ \exists a_j \in A \text{ such that } b_i = f(a_j)$$

Could someone explain the inverse of a bijection, to prove it is a surjection please?

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  • $\begingroup$ You can use LaTeX here. $\endgroup$ – Carsten S Jan 21 '14 at 22:10
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Being a surjection just means you reach all the elements of your target set, here $A$. It is quite easy to show it here: take $x\in A$, then $x=f^{-1}(f(x))$, so $x$ is reached by $f^{-1}$. Therefore $f^{-1}$ is surjective.

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It's basically the definition of a function. Let $a\in A$ and suppose $f(a)=b$. By definition $f^{-1}(b)=a$ and so for every $y\in A$ there exists an $x\in B$ such that $f^{-1}(x)=y$. Hence $f^{-1}$ is surjective.

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For all of the following $a \in A, b \in B$.

$f:A \to B$ and it's inverse $f^{-1}:B \to A$

Since $f^{-1}$ is the inverse of $f$ we know that if $(a,b)\in f$ then $(b,a)\in f^{-1}$.

Assume that $f^{-1}$ is not injective then $\exists (b_1,a),(b_2,a)\in f^{-1}$ which implies that $\exists (a,b_1),(a,b_2)\in f$ which means $f$ is not injective. a contradiction. Therefore $f^{-1}$ is injective.

Now assume $f^{-1}$ is not surjective then $\exists a \in A$ s.t. $(b,a)\notin f^{-1}$. From the definition of inverse this means $\exists a \in A$ s.t. $(a,b) \notin f$ which is again a contradiction. Therefore $f^{-1}$ is a surjection.

Therefore $f^{-1}$ is a bijection.

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